on the denseness of in

Journal of Functional Analysis 238 (2006) 463–470

www.elsevier.com/locate/jfa

On the denseness of span{xλj (1 − x)1−λj } in C0([0,1])Tamás Erdélyi

Department of Mathematics, Texas A&M University, College Station, TX 77843, USA

Received 9 October 2005; accepted 27 January 2006

Available online 4 April 2006

Communicated by G. Pisier

Abstract

Associated with a sequence Λ = (λj )∞j=0 of distinct exponents λj ∈ [0,1], we define

H(Λ) := span{xλ0(1 − x)1−λ0 , xλ1(1 − x)1−λ1 , . . .

} ⊂ C([0,1]).

Answering a question of Giuseppe Mastroianni, we show that H(Λ) is dense in

C0[0,1] := {f ∈ C[0,1]: f (0) = f (1) = 0

}in the uniform norm on [0,1] if and only if

∞∑j=0

(1/2 − |1/2 − λj |) = ∞.

© 2006 Elsevier Inc. All rights reserved.

MSC: 41A17

Keywords: Density of function spaces; Müntz type theorems; Inequalities for Müntz polynomials

Associated with a sequence Λ = (λj )∞j=0 of distinct exponents λj ∈ [0,1], we define

Hn(Λ) := span{xλ0(1 − x)1−λ0 , xλ1(1 − x)1−λ1 , . . . , xλn(1 − x)1−λn

} ⊂ C([0,1])

E-mail address: [email protected].

0022-1236/$ – see front matter © 2006 Elsevier Inc. All rights reserved.doi:10.1016/j.jfa.2006.01.020

464 T. Erdélyi / Journal of Functional Analysis 238 (2006) 463–470

and

H(Λ) :=∞⋃

n=0

Hn(Λ) = span{xλ0(1 − x)1−λ0 , xλ1(1 − x)1−λ1 , . . .

} ⊂ C([0,1]).

In June, 2005, G. Mastroianni approached me with the following question. Let Λ = (λj )∞j=0 be

an enumeration of the rational numbers in (0,1). Is it true that H(Λ) is dense in C0([0,1]) :={f ∈ C([0,1]): f (0) = f (1) = 0} in the uniform norm on [0,1]? In this note we answer hisquestion by proving the following result.

Theorem. Let Λ = (λj )∞j=0 be a sequence of exponents λj ∈ (0,1). H(Λ) is dense in C0([0,1])

in the uniform norm on [0,1] if and only if

∞∑j=0

(1/2 − |1/2 − λj |

) = ∞. (1)

Throughout the paper we adopt the notation

‖f ‖A := supx∈A

∣∣f (x)∣∣

for complex-valued functions f defined on a set A.To prove the “if part” of the theorem, we need the lemma below. It is a well-known conse-

quence of Jensen’s formula.

Lemma 1. Suppose f is a bounded analytic function on the open disk D(a, r) centered at a

with radius r . If (zj )∞j=0 is a sequence of distinct complex numbers such that zj ∈ D(a, r) and

f (zj ) = 0 for each j = 0,1, . . . , and

∞∑j=0

(r − |zj − a|) = ∞,

then f ≡ 0 on D(a, r).

Proof of the “if part” of the theorem. Suppose H(Λ) is not dense in C0([0,1]) in the uniformnorm on [0,1]. Combining the Hahn–Banach theorem and the Riesz representation theorem, wehave a Borel measure with finite total variation on [0,1] such that

supp(μ) ∩ (0,1) �= ∅,

and

1∫xλj (1 − x)1−λj dμ(x) = 0, j = 0,1, . . . .

0

T. Erdélyi / Journal of Functional Analysis 238 (2006) 463–470 465

Then the function

f (z) :=1∫

0

tz(1 − t)1−z dμ(t)

is a bounded analytic function on the strip

S := {z ∈ C: 0 < Re(z) < 1

},

hence it is a bounded analytic function on the open disk D(1/2,1/2) centered at 1/2 with radius1/2. Since f (λj ) = 0 for each j = 0,1, . . . , and

∞∑j=0

(1/2 − |1/2 − λj |

) = ∞,

Lemma 1 implies that f ≡ 0 on D(1/2,1/2), hence by the unicity theorem f ≡ 0 on the strip S

as well. Hence

f (1/2 + iy) =1∫

0

t1/2+iy(1 − t)1/2−iy dμ(t) = 0

for every y ∈ R. Therefore, for every y ∈ R,

0 = f (1/2 + iy) =1∫

0

√t (1 − t)

(t

1 − t

)iy

dμ(t) =∞∫

−∞eixy dν(x),

with a non-zero Borel measure ν with finite total variation on R. However, this is a contradiction,see the exercise at the end of Section 2.1 of [7, Chapter VII], for instance. �Lemma 2. We have

∣∣y(1 − y)Q′(y)∣∣ �

(1 + 9

n∑j=0

λj

)‖Q‖[0,1]

for every Q ∈ Hn(Λ) and y ∈ (0,1).

Associated with a sequence Λ = (λj )∞j=0 of distinct nonnegative exponents λj , we define

Mn(Λ) := span{xλ0 , xλ1, . . . , xλn

}.

To prove Lemma 2, we need D.J. Newman’s inequality [2, Theorem 6.1.1, p. 276].


Lemma 3. Let b > 0. The inequality

∣∣yP ′(y)∣∣ � 9

(n∑

j=0

λj

)‖P ‖[0,b]

holds for every P ∈ Mn(Λ) and y ∈ (0, b].

Note that D.J. Newman [5] proves the above inequality with the constant 11 rather than 9.

Proof of Lemma 2. Note that every Q ∈ Hn(Λ) is of the form

Q(x) = (1 − x)P

(x

1 − x

)(2)

with some P ∈ Mn(Λ). Let y ∈ (0,1). Using Lemma 3, we obtain

∣∣y(1 − y)Q′(y)∣∣ =

∣∣∣∣y(1 − y)(1 − y)P ′(

y

1 − y

)1

(1 − y)2− y(1 − y)P

(y

1 − y

)∣∣∣∣�

∣∣∣∣yP ′(

y

1 − y

)∣∣∣∣ +∣∣∣∣(1 − y)P

(y

1 − y

)∣∣∣∣� y

1 − y

y9

(n∑

j=0

λj

)‖P ‖[0,

y1−y

] +∣∣∣∣(1 − y)P

(y

1 − y

)∣∣∣∣� 9

(n∑

j=0

λj

)(1 − y)

∥∥∥∥P

(u

1 − u

)∥∥∥∥[0,y]+ ∣∣Q(y)

∣∣

� 9

(n∑

j=0

λj

)∥∥∥∥(1 − u)P

(u

1 − u

)∥∥∥∥[0,y]+ ∣∣Q(y)

∣∣

�(

1 + 9n∑

j=0

λj

)‖Q‖[0,y] �

(1 + 9

n∑j=0

λj

)‖Q‖[0,1]. �

We need the following version of a simple lemma from [4], the short proof of which is pre-sented in this note as well.

Lemma 4. Let Γ := (γj )∞j=0 be a sequence of distinct nonnegative real numbers with

η :=∞∑

j=0

γj < ∞.

Then

∣∣P(z)∣∣ � exp

(9η‖ log z‖K

)‖P ‖[0,1], z ∈ K,


for every P ∈ M(Γ ) := span{xγ0, xγ1 , . . .} and for every compact K ⊂ C \ {0}.

In fact, in the “only if part” of the proof of the theorem, the following consequence ofLemma 4 is needed.

Lemma 5. Let Γ := (γj )∞j=0 be a sequence of distinct nonnegative real numbers with

η :=∞∑

j=0

γj � 1

20.

Then

‖Q‖[3/4,1] < ‖Q‖[0,1/2]

for every Q ∈ H(Γ ) = span{xγ0(1 − x)1−γ0 , xγ1(1 − x)1−γ1 , . . .}.

Proof of Lemma 4. The lemma is a consequence of D.J. Newman’s Markov-type inequality.Repeated applications of Lemma 3 with b := 1 and the substitution x = e−t imply that

∥∥(P

(e−t

))(m)∥∥[0,∞)� (9η)m

∥∥P(e−t

)∥∥[0,∞), m = 1,2, . . . ,

and, in particular

∣∣(P (e−t

))(m)(0)

∣∣ � (9η)m∥∥P(e−t )

∥∥[0,∞), m = 1,2, . . . ,

for every P ∈ M(Γ ). By using the Taylor series expansion of P(e−t ) around 0, we obtain that∣∣P(z)∣∣ � c1(K,η)‖P ‖[0,1], z ∈ K,

for every P ∈ M(Γ ) = span{xγ0 , xγ1, . . .} and for every compact K ⊂ C \ {0}, where

c1(K,η) :=∞∑

m=0

(9η)m‖log z‖mK

m! = exp(9η‖log z‖K

),

and the result of the lemma follows. �Proof of Lemma 5. This follows from Lemma 4 and the fact that every

Q ∈ Hn(Γ ) = span{xγ0(1 − x)1−γ0 , xγ1(1 − x)1−γ1 , . . . , xγn(1 − x)1−γn

}(3)

is of the form (2) with a

P ∈ Mn(Γ ) = span{xγ0, xγ1 , . . . , xγn

}. � (4)

Let Γ := (γj )∞j=0 be a sequence of distinct nonnegative real numbers with γ0 := 0. One of the

most basic properties of a Müntz space Mn(Γ ) defined in (4) is the fact that it is a Chebyshev


space of dimension n+1 on every A ⊂ [0,∞) containing at least n+1 points. That is, Mn(Γ ) ⊂C(A) and every P ∈ Mn(Γ ) having at least n + 1 (distinct) zeros in A is identically 0. Since anyQ ∈ Hn(Γ ) defined in (3) is of the form (2) with a P ∈ Mn(Γ ) defined in (4), the space Hn(Γ ) isalso a Chebyshev space of dimension n + 1 on every A ⊂ [0,1) containing at least n + 1 points.The following properties of the space Hn(Γ ), as a Chebyshev space of dimension n+ 1 on everyA ⊂ [0,1) containing at least n + 1 points, are well known (see, for example, [2,3,6]).

Lemma 6 (Existence of Chebyshev polynomials). Let A be a compact subset of [0,1) containingat least n + 1 points. Then there exists a unique (extended) Chebyshev polynomial

Tn := Tn{γ0, γ1, . . . , γn;A}

for Hn(Γ ) on A defined by

Tn(x) = c

(xγn(1 − x)1−γn −

n−1∑j=0

ajxγj (1 − x)1−γj

),

where the numbers a0, a1, . . . , an−1 ∈ R are chosen to minimize

∥∥∥∥∥xγn(1 − x)1−γn −n−1∑j=0

ajxγj (1 − x)1−γj

∥∥∥∥∥A

and where c ∈ R is a normalization constant chosen so that

‖Tn‖A = 1

and the sign of c is determined by

Tn(maxA) > 0.

Lemma 7 (Alternation characterization). The Chebyshev polynomial

Tn := Tn{γ0, γ1, . . . , γn;A} ∈ Hn(Γ )

is uniquely characterized by the existence of an alternation set

{x0 < x1 < · · · < xn} ⊂ A

for which

Tn(xj ) = (−1)n−j = (−1)n−j‖Tn‖A, j = 0,1, . . . , n.

Note that the existence of a unique (extended) Chebyshev polynomial with the above alterna-tion characterization can be guaranteed even if we allow that A = [0,1] rather than A ⊂ [0,1).This can be seen by a standard limiting argument. Namely we take the Chebyshev polynomialsTn,δ on the interval [0,1 − δ] first and then we let δ > 0 tend to 0.


Now we are ready to prove the “only if part” of the theorem. It turns out that the approachused in the corresponding part of the proof in [1] can be followed here, but, while the proof isstill rather short, the details are slightly more subtle.

Proof of the “only if part” of the theorem. Suppose now that Λ is a sequence of distinctexponents λj ∈ (0,1) such that (1) does not hold, that is,

∞∑j=0

(1/2 − |1/2 − λj |

)< ∞.

Then there are sequences Γ := (γj )∞j=0 of distinct numbers γj ∈ [0,1/4) and Δ := (δj )

∞j=0

of distinct numbers δj ∈ [0,1/4), and a finite set {α1, α2, . . . , αm} of distinct numbers αj ∈[1/4,3/4] such that

∞∑j=0

γj <1

20,

∞∑j=0

δj <1

20,

and

{λj : j = 0,1, . . .} ⊂ {γj : j = 0,1, . . .} ∪ {1 − δj : j = 0,1, . . .} ∪ {αj : j = 1,2, . . . ,m}.

Without loss of generality we may assume that γ0 := 0 and δ0 := 0. For notational convenience,let

Tn,γ := Tn

{γ0, γ1, . . . , γn; [0,1]},

Tn,δ := Tn

{1 − δ0,1 − δ1, . . . ,1 − δn; [0,1]},

T2n+m+1,γ,δ,α := T2n+m+1{γ0, . . . , γn,1 − δ0, . . . ,1 − δn,α1, . . . , αm; [0,1]}.

It follows from Lemmas 2, 5, and 7, and the Mean Value theorem that for every ε > 0 there existsa k1(ε) ∈ N depending only on (γj )

∞j=0 and ε (and not on n) so that Tn,γ has at most k1(ε) zeros

in [ε,1] and at least n − k1(ε) zeros in (0, ε). Similarly, applying Lemmas 2, 5, and 7, and theMean Value theorem to Sn,δ(x) := Tn,δ(1 − x) gives that for every ε > 0 there exists a k2(ε) ∈ N

depending only on (δj )∞j=0 and ε (and not on n) so that Tn,δ has at most k2(ε) zeros in [0,1 − ε]

and at least n − k2(ε) zeros (1 − ε,1).Now, counting the zeros of Tn,γ − T2n+m+1,γ,δ,α and Tn,δ − T2n+m+1,γ,δ,α , we can deduce

that T2n+m+1,γ,δ,α has at least n− k1(ε)− 3 zeros in (0, ε) and it has at least n− k2(ε)− 3 zerosin (1− ε,1) (we count every zero without sign change twice). Hence, for every ε > 0 there existsa k(ε) ∈ N depending only on (λj )

∞j=0 and ε (and not on n) so that T2n+m+1,γ,δ,α has at most

k(ε) zeros in [ε,1 − ε].Let ε := 1/4 and k := k(1/4). Pick k + m + 5 points

1< η0 < η1 < · · · < ηk+m+4 <

3

4 4


and a function f ∈ C0([0,1]) so that f (x) = 0 for all x ∈ [0,1/4] ∪ [3/4,1], while

f (ηj ) := 2(−1)j , j = 0,1, . . . , k + m + 4.

Assume that there exists a Q ∈ H(Λ) so that

‖f − Q‖[0,1] < 1.

Then Q − T2n+m+1,γ,δ,α has at least 2n + m + 2 zeros in (0,1). However, for sufficiently largen, Q − T2n+m+1,γ,δ,α is in the linear span of the 2n + m + 2 functions

xγj (1 − x)1−γj , j = 0,1, . . . , n,

x1−δj (1 − x)δj , j = 0,1, . . . , n,

and

xαj (1 − x)1−αj , j = 1,2, . . . ,m,

so it can have at most 2n + m + 1 zeros in (0,1). This contradiction shows that H(Λ) is notdense in C0([0,1]). �References

[1] P.B. Borwein, T. Erdélyi, The full Müntz theorem in C[0,1] and L1[0,1], J. London Math. Soc. 54 (1996) 102–110.[2] P.B. Borwein, T. Erdélyi, Polynomials and Polynomials Inequalities, Springer-Verlag, New York, 1995.[3] E.W. Cheney, Introduction to Approximation Theory, McGraw–Hill, New York, 1966.[4] T. Erdélyi, W. Johnson, The “Full Müntz Theorem” in Lp[0,1] for 0 < p < ∞, J. Anal. Math. 84 (2001) 145–172.[5] D.J. Newman, Derivative bounds for Müntz polynomials, J. Approx. Theory 18 (1976) 360–362.[6] T.J. Rivlin, Chebyshev Polynomials, second ed., Wiley, New York, 1990.[7] B. Sz.-Nagy, Introduction to Real Functions and Orthogonal Expansions, Oxford Univ. Press, New York, 1965.

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